Let me ask a question which appears in the book 'Elementary Differential Geometry' written by O'Neil.
The questions is: prove that if a one-to-one and onto mapping $f:\Bbb R ^n \to \Bbb R ^n$ is regular, then it is diffeomorphism.
In the book, "$f$ is regular" means that the tangent map of $f$ is one to one. I think certainly I need to use the inverse function thorem. $f$ is a mapping so $f$ is in the class $\mathcal C ^1$. Since $f$ is regular and in the class $\mathcal C^1$, its Jacobian is invertible, so the derivative of $f$ is invertible.
Therefore, I might apply the inverse function theorem to $f$: there exists an open set in $\Bbb R ^n$ in which the inverse of $f$ exists and the inverse belongs to the class $\mathcal C ^1$ so $f$ is a diffeomorphism. But to prove $f$ is a diffeomorphism, shouldn't I show the inverse of $f$ is in the class $\mathcal C ^1$ for every point in $\Bbb R ^n$?
I would really appreciate any help. Thank you for reading.
Best Answer
Because $f$ is bijective, it has an inverse. Your inverse function theorem argument shows $f^{-1}$ is "locally $C^{1}$" at an arbitrary point (though you might want to supply a bit more detail to show the "arbitrary point" part). But being $C^{1}$ is a local condition, so you've actually shown $f^{-1}$ is $C^{1}$.
On a tangent, the fact that $f$ maps $\mathbf{R}^{n}$ to $\mathbf{R}^{n}$ (a space of the same dimension) is crucial: It's not generally true that a regular injection is a diffeomorphism onto its image. The plane curve $c:(-\pi, \pi) \to \mathbf{R}^{2}$ defined by $c(t) = (\sin 2t, \sin t)$ is regular and injective, but its image is a figure-8, which is not diffeomorphic to an interval.